JPJC 2026 Oscillations Assignment Suggested Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS OSCILLATIONS ASSIGNMENT SOLUTIONS 1 (a) (i) From the graph, T = 0.60 s. [1] (ii) The angular frequency, T 2 60.02 [1] 110.5 rad s  [1] (b) (i) From the graph, the time interval is 0.20 s. [1] (ii) Phase angle  2Tt  260.0 20.0 [1] 09.2 radians [1] (c) (i) Light damping of the oscillations may be achieved by sticking a thin, flat card to the mass without adding significant mass. This increases the surface area of the mass exposed to air to enhance air damping. [1] (ii) The degree of damping may be increased by immersing the mass into a liquid that is more viscous, such as oil. [1] 2 Since the mass starts its motion at maximum negative displacement (taking direction upwards as positive), then 0 cosx x t  . (a) 0 cosx x t        0.10.22cos0.3x 0.3x cm [1] The mass will be at a distance of 3.0 cm above its equilibrium position. Hence, the total distance the mass has moved is 3.0 + 3.0 = 6.0 cm. [1] (b) 0 cosx x t  23.0 cos 0.452.0x       0.47 cmx  The mass will be at a distance of 0.47 cm below its equilibrium position. [1] Hence, the total distance the mass has moved is 3.0 + 0.47 = 2.5 cm. [1]

2 3 (a) (i) The graph is a straight line that passes through the origin; this means that the acceleration is directly proportional to the displacement. The negative gradient implies that a and x are always in opposite direction. Hence, this shows a is always directed towards the equilibrium. Therefore, the motion can be represented by 2a x   and is simple harmonic. [2] (ii) 2Using ,a x   2 2.20 2.20gradient 55 0.040 ( 0.040)        [1] 2 2 2 2 4 4 0.85 s55 T T       [1] (b) 2 21Using 2max oKE m x      2 2 2 1 2 1 0 25 55 0 042 max oKE m x . .   [1] 0 011J. [1] (c) [1] for correct shape [1] for correct labelling Ek / J 0 T 2T t 0.011